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How heading bug is calculated during cross wind and tail wind?
From this question, it is been suggested to ask different question, rather than keep discussing it under comment section.

How is the heading bug calculated and used for navigation when there is a cross wind?
… when there is a tail wind?
Will the heading bug matter for navigation when there is wind correction required?
What role does the heading bug play in autopilot mode?

Possible duplicate of What are the differences between Bearing vs Course vs Direction vs Heading vs Track?

I don't think the linked question is similar to this one at all. They just happen to mention the same terms. Certainly none of the answers on the linked question is a useful answer to this question.

First thing first, let's go over some concept: heading is which way the plane's nose is pointing at, track is which way the plane is travelling on the ground. E.g. if your heading is 360 (North), and the wind is blowing from 090 (blowing East to West), the plane's track may be 350 (slightly left of North).
With that in mind, it's not hard to deduce most of the time pilots want to fly track, rather than heading. The purpose is to navigate from point A to point B on the ground. In our example, you have to turn the plane slightly to the right to compensate for the crosswind. To figure out exactly how many degrees to turn, you will first need to decide how fast you want to fly. After that, it's simple geometry.

A mechanical flight computer can be used to calculate the drift angle easily. Using it is a matter of marking dots on a transparent disc, then sliding and rotating it to the correct position and read off the pre-calculated figures below.

If you are flying manually, or flying a plane with an elementary autopilot, you will need to calculate the drift angle, then add (or minus) that to your desired track to get the heading. You will then set the heading bug to that value. Theoretically, that heading should compensate just enough crosswind that you stay on your desired ground track. In practice minor adjustments are often necessary. Provided that the numbers you use are not awfully off, you should end up with an estimation good enough for the safety of your flight.
If you are flying a plane with an advanced flight computer which has the ability to detect winds aloft, your autopilot will fly the track you dial in and automatically compensate for crosswind.

So to answer your questions:

You will use simple math or tools to compute the heading which will let you fly the desired track.
It does not matter when you have an exact headwind or tailwind.
Heading bug denotes the desired heading, not track. You always turn your plane towards the heading bug.
Advanced autopilot fly track, fundamental ones fly heading. Some autopilots can read the heading bug position on the instrument, others must be set explicitly on another panel.

For cross-wind corrections: on the ground you can use a nice software or even an ol' CRP to compute a drift. For setting the heading bug though, and for computing this in the cockpit, you need a quick way of doing it in your head: one great method involves the 'watch face' :

A) Divide the wind speed (in knots) by your NPM( nautical miles per minute). This is your Max Drift.
B) Use the watch face to determine how much of that Max Drift you need to use: 15 is a quarter of an hour, so for 15 degrees you use 1/4, 30 degrees you use 1/2 of Max Drift, 45 deg 3/4 and for larger degrees use the hole Max Drift
eg. wind 230/45 , your speed 180 kts (or 3 miles per minute), hdg 270

   Max Drift = 45 / 3 = 15 degrees
   Angle between wind and me : 40 degrees, so I use a little more than 1/2 and less than 3/4
  (between 30 and 45 on the watch face)
   Computed Drift = 10 degrees
   Set Hdg Bug to 260 to keep track 270

Hdg bug will not be affected by tail wind, but your ground speed and ETA will be affected. A CRP or an apropriate app  sheet can tell you by how much.
It will matter if a) wind is very strong or b) your navigation leg is long enough (30+ mins)
Depending on the sophistication of the autopilot, most of them have a "keep heading mode" where the a/c will maintain the heading selected by the heading bug. If operated in that mode, the pilot can turn the a/c by moving the hdg knob left or right.

This answer is hard to understand. 1) What is CRP?  2) When you first start talking about "15 degrees", you don't say what are the two things that at a 15-degree angle to each other.  You mean the wind is at a 15-degree angle to the desired ground track?  3) "It" will matter-- what will matter?  You mean in the cases that follow, the tailwind WILL have an effect on the required setting of the heading bug?  If that's what you mean it's not all that clear.

To track a course in a crosswind using the bug, set the OBS to the waypoint you wish to fly to and use your normal crabbing skills to track the needle (025 below). Once the needle is being tracked, you can set the heading bug to the current heading (015) and just fly that. This works with or without an autopilot.

sorry, this is the best picture i could find :(

That assumes you have a radio navigation aid, like VOR or GPS.


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Does FAA simulator (AATD/BATD) count toward experimental rating 1,000 hours requirement?
In the USA, a pilot can obtain an AEA (authorized experimental aircraft) rating to fly demilled surplus military turbojet-powered aircraft under FAA 5-1580(A).  Part of its requirements are to have logged a minimum of 1,000 hours pilot flight time, including 500 hours as pilot in charge.
A student pilot can NOT log hours in a FAA-aproved AATD or BATD simulator, unless an instructor is present.
But, AFTER someone has their private pilot's license, are they then able to log some hours in an FAA-approved AATD or BATD solo (without instructor) toward their 1,000 and/or 500 hours required for the experimental rating?  (Perhaps even using a simulation of the surplus military aircraft they want to have authorization for?)  Or, is that level of distrust from un-verifiability still there after someone has their private pilot's license?

I don't have specifics for this case, but in general sim / FTD time is never useful at any certificate level unless you have an instructor providing instruction.

@casey, I could be misunderstanding, but I posted this under the impression that certain simulated hours for an airline transport pilot can be counted in an FAA-approved simulator, without an instructor.  For an ATP, 1500 hours are required, 100 of which can be in a simulator.  (There's a lot of details about how these hours are broken up, which I'm leaving out.)  Didn't see that an instructor was required for these, such as simulator hours for an instrument rating.  This (ATP) led to my hope for the experimental rating.

The short answer is no. As you said in your question, the regulations you quoted require you to have logged flight time, which is defined in 14 CFR 1.1:

Flight time means:
(1) Pilot time that commences when an aircraft moves under its own
  power for the purpose of flight and ends when the aircraft comes to
  rest after landing; or

In other words, you have to go fly a real aircraft. And 1.1 defines "aircraft" clearly too, so you can't argue that it includes simulators.
But even if you could use a simulator, you would still need an instructor present. 14 CFR 61.51 talks about logging simulator time for various purposes, but in every case it also says that an "authorized instructor" must be present and endorse your logbook. That makes sense: if you're playing around by yourself in a simulator, you could be doing anything at all, including reinforcing your own bad habits. The instructor is there to ensure that what you're doing is useful and relevant, and you're learning from it.

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If the efficiency of a turbofan engine is 35%, where does the rest of the fuel energy go?
There have been a couple of questions on this site about the efficiency and propulsive power of turbine engines (here, here, here).

In a high bypass turbofan engine, what are the losses in the different stages? Where does the unutilised energy go, and what exactly is left over to make the aeroplane fly?

There are a couple of sources of loss throughout the process, as indicated in the figure from an old paper format uni book. I've had to translate the labelling, open to suggestions there. The percentages are valid for a high bypass turbofan manufactured in the late 80s.

The total energy input starts with the fuel flow: chemical energy per second.
Combustion converts the chemical energy into a heat flow pretty successfully, with about 1% lost in incomplete combustion. This takes place in the combustion chamber.
The turbine extracts mechanical energy from the heat flow, and uses a portion of it to power the compressor. The thermodynamic efficiency of the Carnot cycle determines the resulting power fraction. I've labelled the resulting net mechanical power as Gas Power which sounds a bit dicey. This Gas Power can be converted into shaft power and/or into jet power, depending on the type of turbine engine.
Note that the thermodynamic efficiency depends on the inflow speed of the air into the combustion chamber: it is decelerated and compressed in the intake, allowing for higher pressure ratios which result in higher efficiencies.
We now need to utilise the Gas Power to raise the kinetic energy of the medium that is used for propulsion. This increase in kinetic energy (from the reference frame of the aeroplane) is labeled Propulsive Power.
Aero engines can be divided into two major groups:

a. Transformation of available Gas Power into shaft power, delivering mechanical energy which can drive a propeller or rotor, which then increases the kinetic energy of a mass flow of surrounding air - a turboshaft.
b. Direct transformation of available Gas Power into kinetic energy by expansion in a nozzle - a turbojet.

The principal difference between the two is that in a. the mass flow  delivering the propulsion is larger than the mass flow through the turbine, while in b. the mass flow through the turbine is equal to the propulsion mass flow. Since thrust T = $\dot{m} \cdot \Delta V$, this means that at a given thrust and entry velocity, the exhaust velocity of a. must be lower than of b. Note that both the turboprop and the turbofan are a mix of a. and b. since a portion of the turbine mass flow is expanded for jet propulsion.
Converting Gas Power into Propulsion Power cannot take place isentropically (loss free): in a turbofan about 5% is lost into heat flow, less in a turboprop.
The propulsive power is the power transferred to the medium used for propulsion (air or combustion gas). The Propelling Power is the power transferred to the aeroplane. There are efficiency losses in this process as well: the medium exits the propulsion assembly at a higher velocity than the airspeed of the aeroplane, and therefore has a certain absolute velocity (relative to earth). The corresponding kinetic energy flow must be considered as a loss. Of course, the outflow velocity must be greater than the airspeed in order to generate thrust.

This power transformation is therefore also associated with an efficiency factor, $\eta$ propulsion. It costs power to drive those fan blades, which have induced drag and profile drag just like a wing does.

When using these definitions, we can see that an aeroplane on the runway, just before take-off, with the brakes applied and the throttle fully open, has:

Maximum propulsive power since the airflow has maximum $\Delta V$. As computed in this answer.
Zero propelling power, since none of the propulsive power has been transferred to the aeroplane yet. All propulsive power is transformed into kinetic energy of the gas flow.

So if an engine has a static thrust rating of xxx lbs, is that really referring to "propulsive power" and not "thrust" per se?  If an engine is running and producing 2000 lbs of static thrust, with the engine stationary, tied to a very strong fish scale, which will show 2000 lbs, isn't that thrust power?  I think most of us conflate the two terms and I'm having trouble separating them conceptually.

@JohnK Yes it might be a confusing term, like said I am open to suggestions. Static thrust is a force - multiplied by the aeroplane velocity it becomes the power that is now labeled Thrust Power. Propulsive power would be defined by the mass flow of air through the propeller = $\frac{1}{2} \dot{m} V^2$

@JohnK Have changed Thrust Power into Propelling Power, a fan is a kind of propeller after all.

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How does flying a helicopter compare with flying a fixed wing plane?
There has been a lot of interest in the ease of flying a helicopter recently, and questions have arisen on how easy it is to control a helicopter in less than ideal circumstances such as low visibility.
What are the main difficulties in controlling a helicopter, compared to controlling a fixed wing aeroplane in similar circumstances?

Controlling a helicopter is trickier than controlling a fixed wing aeroplane, because:

a helicopter can hover, in which case it is unstable.
the helicopter rotor is horizontal,
the rotor has long blades, applying a lot of torque to the fuselage.

The hover. in a conventional helicopter with a single rotor on top, the hover is an unstable situation requiring constant correcting inputs with the cyclic stick, as shown in the above clip. The guest pilot on the left seat has a successful go at controlling the cyclic stick (the stick at the centre), while the impressively capable right seat pilot controls the pedals and the collective stick. The helicopter wants to fall away to the sides and forwards/aft all the time, from disturbances caused by the wake interference with the ground, wind etc. Random accelerations need to be corrected by providing control corrections, based on what the pilot sees in relation with the environment. Peripheral vision is pretty crucial in this, since it rapidly and accurately picks up movements. Four degrees-of-freedom need to be actively controlled: pitch, roll, yaw, vertical speed.
That is all to hover at constant altitude. Now hover and climb:

increase collective (more eying power), which results in more torque;
so compensate with the pedal input for the tail rotor, which increases sideways thrust;
which must be compensated by lateral cyclic. Three control inputs for one degree-of-freedom!

Horizontal rotor. Once the helicopter picks up speed, its vertical and horizontal stabiliser keep the nose more or less aligned into the airflow - more at higher speed, but the neutral angle also changes with speed. The picture above is also shown in this answer, and depicts what happens with relative blade speed in forward flight. The blade going forward creates more lift than the blade going aft (which is compensated by increasing the Angle of Attack (AoA) of the retreating blade), but there is an area of reverse flow to the side of the rotor axis which results in a sideways tilt of the rotor. Which must be compensated by a cyclic stick left input, more stick at higher speed.
Rotor torque. Installed engine power equates to torque applied to the rotor, times the rotor rotational speed. A propeller is smaller than a rotor and rotates faster: the propeller requires less torque application for a given engine power setting, and the ailerons at the end of the long fixed wings can easily compensate for that. The helicopter rotor requires a lot of torque, which requires a tail rotor to compensate. The faster the helicopter flies, the more torque, the more force required from the tail rotor - which thrusts sideways, creating a drift in the flight path, which must be trimmed by the lateral stick. Even with a vertical stabiliser, tail rotor thrust is a function of airspeed.

Humans can learn to control helicopters of course, but there is a lot more to it than controlling a fixed wing aeroplane. Our brains rely on inner ear and visual inputs for detecting accelerations. More brain stimulation by input detection via multiple channels, results in more precise control. Plus experience of course.But take away an input, like the peripheral vision when flying in reduced visibility, and the control of a helicopter becomes much harder.
In larger helicopters such as S76s, pilot control is helped by a Stability Augmentation System (SAS), an Inner Loop system with no feedback on the flight controls; and an Automatic Flight Control System (AFCS) which changes the trim point of the flight controls as a function of flight path and target point. But these systems are not flight critical: if they fail, the pilot must take over, and must be able to fully control the helicopter by manual input. From visual and inner ear cues, and this always remains harder than in the auto stabilising fixed wing.
Fixed wing aeroplanes can and must be designed to be aerodynamically stable. Helicopters cannot.

I had a chance to fly a Robinson R-66 a couple of years ago and can say that in the hover it is almost exactly like balancing a broom stick on your hand, while holding a cup of coffee with the other and nudging a soccer ball along the ground with your feet.  In forward flight is was like a neutrally stable and very sensitive and twitchy fixed wing airplane.

"Fixed wing aeroplanes can and must be designed to be aerodynamically stable. Helicopters cannot".  Hmmm.   A) actually not all fixed-wing airplanes are designed to be aerodynamically stable, examples being computer-assisted FBW aircraft and even some neutrally stable (or slightly negatively st?) full-scale and miniature (rc) aerobatic sport planes  and B) if a helicopter had a tall enough rotor mast and counter-rotating blades, would it not be aerodynamically stable?  I've seen plans for a free-flight (no pilot input of any kind) rubber-band powered model helicopter designed along such lines.

The stability is due in part to some version of the "pendulum effect", of course.  :)

When I had the chance to fly a Kamov-28, I found it stable and easy to control, even in hover. But that is only thanks to the coaxial rotor system.

@quietflyer, only FBW fighter aircraft are aerodynamically unstable, but artificially still neutral and if the augmentation fails (it does have at least double backup), there is the eject option. All civilian FBW aircraft are aerodynamically stable, though some are then artificially neutral. But that just mean they maintain flight path angle instead of speed with pitch. They maintain flight path angle quite well.

@JanHudec -- every example in my comment AFTER the phrase "computer-assisted FBW aircraft" was intended to refer specifically to NON-FBW aircraft.  I should have been more clear.  I think some civilian aerobatic aircraft are routinely flown at CG locations where pitch stability is neutral or slightly negative.  I may be wrong.  Certainly this seems to the case with some RC models.  And there's no question that roll stability is neutral or negative in some civilian a/c.  E.g. some sailplanes.

@quietflyer, that does not affect disputing the part before that in any way. Also, are you sure any aerobatic (full-scale—anybody can do what they want with rc) are unstable? Because I doubt it would be certifiable.

The first poster is spot on. Additionally, the most common response to emergencies is the exact opposite between helicopters and airplanes.
The first reaction ingrained in you for emergencies in an airplane is to push the yoke forward to lower the angle of attack, and to open the throttle to full (feather the inop engine and slightly retard the throttle of the good engine in the case of an engine failure in a multi-engine).
The first reaction ingrained in you for emergencies in a helicopter is to pull the cyclical (yoke) back, and to lower the collective.
I don’t have a lot of hours in helicopters. When people ask me what it feels like to me this is my response.

Flying an airplane is like driving a car. It is inherently stable. You can fly it straight, level and unaccelerated hands off as long as you pay attention.
Flying a helicopter is more like riding a motorcycle. It is inherently statically unstable. Your entire body is fully involved at all times. Your attention should also be fully involved.

I used to ride motorcycles and I'd quibble with that one.  You can let go of the handle bars on a motorcycle, even stand on it, while it's at speed.  Riding a motorcycle at idle in first gear, well maybe yeah.

LOL. Yes, but any idiot can drive a car and let go of the steering wheel. It takes a very special one to take their hands off the handlebars of a motorcycle.

Seriously though, it takes much more skill and coordination to fly a helicopter than an airplane in normal flight, IMHO. Same with motorcycles and cars. I’ve done all four. Aerobatics and stunting takes even more skill and coordination. A helicopter instructor and DPE working with the US Helicopter Safety Team once told me that the same pilots who ride motorcycles are the same ones who would fly helicopters. The mentality and personality types are the same.

It's an excellent analogy.  And John's addendum ... motorcycle at idle in first gear is an even better analogy!

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Have there been fully pneumatic aircraft?
Have there been any passenger aircraft that used pneumatics for all of their on board systems that required force actuation, such as:

landing gear;
wheel brakes;
nose wheel steering;
propeller brakes;
passenger door operation?

Yes there have been, and some of them are still flying.

This article is about the fully pneumatic actuation systems on board the F27s built by Fokker and by Fairchild. If there was a leak, it was leaking clean air. System pressure was 3,300 PSI = 228 bar.
Some of the advantages cited:

Many factors favored the use of pneumatic systems on the F-27, most significant of which was an approximate 100-lb weight saving over electrical and hydraulic systems.

Lighter weight
Faster operation
No flammable liquids
Cleaner and easier to maintain

The systems were designed 50 years ago, fully aeronautical lightweight design. Could most definitely be used to pump up the tyres in this question to 16 bar.

From the source it is clear that compressed air powered landing gear, wheel brakes, nose wheel steering, propeller brakes, and passenger door on this plane. Not the control surfaces, however.

No the control surfaces were fully reversible. The good old days, when pilots of airliners could feel the air flowing past their surfaces. Sigh..

Never realized the F-27 had such tiny horizontal stabs

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Is there any way we can predict an air crash?
Is there any way we can predict a crash from the system inputs?
The question arises from this discussion.

There are solutions for anything if you have no price constraint.

In addition to the systems listed by RedGrittyBrick, it is also possible to let the aircraft detect itself, from its sensors, potential crashes.
This problem has been studied, and is pretty well documented in a report made by the BEA (the French equivalent of the NTSB) after the crash of the flight AF447. The full report is available in French and in English. I made a summary of their work in this answer.
The solution proposed in this report is a triggered transmission of flight data. The aircraft transmits them only if it detects a potentially dangerous situation from its sensors (this reduces the load on communication systems).
Two types of triggering criterion are proposed :

One base on binary logic
One based on Fuzzy logic

They both take into account several parameters (speed, pitch angle, pilot's inputs...) and verify some relations about them. For example, if roll > 50° for more than 2s, something is probably wrong.
The criterion have been check against a database of real flight data recovered from crashed aircrafts, in order to assess their performances :

The robustness of the triggering criteria was evaluated using the following metrics:

False detection rate (or nuisance trigger rate): out of all the normal flights, how many are considered as containing an emergency situation

Nuisance transmission time that the nuisance triggers generate

Emergency situation detection rate: out of all the accident/incident flights, how many are considered as containing an emergency situation

For flights with a correctly detected emergency situation, what is the warning time between the time of detection and the time of impact with the surface

The conclusion of these tests is that accidents were almost always predicted (binary logic failed in only one case out of 44, fuzzy logic detected every case).  On average, the problem was raised about 350s before the actual crash (the median is around 33s, minimum at 3s and maximum at 10,019s).
The second half of the report discusses different system to transmit these triggered alarms to infrastructures on the ground. It is also interesting and well written, but outside the scope of this question.

Yes, many aircraft are fitted with TCAS which predicts one type of crash - an aerial collision.
Many aircraft are fitted with GPWS which predicts another type of crash - controlled flight into terrain.
The ground based ATC have systems that predict a loss of separation, the Short-Term Conflict Alert [STCA]. See this question about separation and conflicts. In terms of predictions this would constitute a very high rate of false positives. But that is good.
Because of the predictions, many crashes are avoided.

+1, but that last sentence explains why these are pretty poor predictors.


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Which are the main aluminium alloys used in airliner construction?
The proportion of composites used in airliner construction has increased in the last decades, with the B787 and the A350 being the first ones with composites making up the majority of wing and fuselage structure. From Wikipedia:

The 787 is the first major commercial airplane to have a composite fuselage, composite wings, and use composites in most other airframe components.

And

The A350 XWB airframe will be made out of 53% composites: carbon fibre reinforced plastic for the outer and centre wing box (covers, stringers, spars), fuselage (skin, frame, keel beam, and rear fuselage) and the empennage (horizontal and vertical tailplanes).

The vast majority of aircraft in operation (A320, A330, B737, B777) still use aluminium as the main load bearing construction material though. Which aluminium alloys were used for the wings and fuselage of these aircraft?

The alloys have different characteristics , so the answers also depend on which particular application.

2000s, 6000s, and 7000s series alloy are very common for aircraft construction over the past 8 decades of metal aircraft design.  Boeing aircraft are largely manufactured from 2000s series aluminum alloy.


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Is the total upwards vertical force (lift included) greater than the weight in a steady climb?

Related to questions on forces in a climb: if any aircraft is climbing with a constant velocity, is the total upwards vertical force (lift included) greater than the weight?

We already have a question that specifically asks "My question therefore is also about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft."  See  Does lift equal weight in a climb?

The body of the question asks about the total upwards force, which might be construed to mean either the net vertical force excluding weight, or the sum of the upward force components.  Since at least one aerodynamic force has a downward component, these are not the same thing.  The title asks about the total vertical force, presumably meaning the net vertical (aerodynamic?) force, i.e. the sum of all vertical forces other than weight. The title and body could use some editing to be better harmonized and be more clear about which of these things you are asking.

(So consider changing title to read "Is the sum of the upwards forces higher than weight in a steady climb?", or "Is the net vertical aerodynamic force greater than weight in a steady climb?", depending on which you really want to know.)

Is Lift Greater Than Weight In A Climb was asked 4 years ago, editing it now may make many of the answers a mismatch. It was active again when this answer was added, initially stating that it was. It received quite some downvotes. It is important to realise in the debate that upwards vertical force is higher than weight in a climb, but lift isn't necessarily, depending on the tilt of the aircraft velocity vector.

It's interesting how vague simple words can end up being-- (I guess that's why a force vector diagram is worth a thousand words) -- I'm still not completely clear whether you are asking about the net vertical aerodynamic force, or the net vertical force, or you are interested in adding up all the upward aerodynamic force components (while ignoring the downward aerodynamic force components) and comparing that value to the weight-- my answer https://aviation.stackexchange.com/a/56476/34686 to the other related question is intended to address all those different cases--

Come on.. F=ma!  If the net vector acceleration is zero, then the sum of all vector forces must also equal zero. What you call them, or how you break them up into arbitrary components is just picking at nits. If you define and separate them in any way where the sum is NOT zero, then you have done it wrong or missed one or more of them. Remember any scheme used to define all the forces is just sleight of hand... The real physical, actual forces are the force exerted by each molecule of air/gas bouncing off the airframe, and the propeller as it spins, or the turbine in the jet engine, etc. etc.

The word Upwards has been added in the title, voting to reopen.

Using a vertical rocket as a model, the answer is obviously yes, as vertical drag + weight = "total upwards vertical force" (in a steady state climb).   What made it so interesting all those years ago was replacing the more efficient wing with thrust to get a higher angle of climb, and comparing that with using thrust to climb faster, but at a lower angle.  Hence the Vx, Vy, cruise climb discussion goes on.

@quietflyer In your answer at Lift Greater Than Weight, you state: “In a steady climb, net vertical force must be zero, so net vertical aerodynamic force must be equal to weight.”. Plus vertical component of drag, as illustrated in the answer.

@Sanchises Are you happy with the changes now?

@Koyovis please draw the closed vector diagram (better yet to scale) that shows thrust that closes the triangle of lift plus thrust that overcomes aerodynamic drag.  Essentially, it's quiet flyers  diagrams plus a "handle on the side" for the equal but opposed drag and thrust vectors.  For all three cases, we can see how excess thrust can be used to climb.  Don't forget greater airspeed makes a bigger "handle", steeper climb can be lower speed, smaller lift vector, but much more use of excess thrust for sin of climb angle.

Also remember (for props) excess thrust is greater at lower airspeeds, so yea, it's complicated. Let's go with 2400 lbs of lift (in level flight), 700 lbs of thrust available at Vx, 650 at Vy, and 550 at cruise climb.  Aerodynamic drag 300 lbs at Vx, 250 lbs at Vy, 350 lbs at cruise climb.  See what we get.  The diagrams will yield angle of climb, then we can throw V as a Power factor (F × V) to get rates of climb.

As sin angle x V.

@RobertDiGiovanni Ehm...that was another question, a couple of them. Point of this one is to acknowledge that not only weight needs to be compensated for by a combination of upward forces. Whichever combination that is, has been answered in a lot of detail..

Yes, lots.  TL/DR?  Closed vectors to scale, haven't seen it, particularly for your question presentation because:  drag at Vx is higher due to less optimal AOA, drag is higher at "cruise" due to less optimal speed.  Vy combines the best of both for highest rate of climb which is greatest excess Power as F × V.  We're both preaching to the choir, but it sounds good.  (and we still have that thrust available curve to think about too).

@Koyovis -- re "@quietflyer In your answer at Lift Greater Than Weight, you state: “In a steady climb, net vertical force must be zero, so net vertical aerodynamic force must be equal to weight.”. Plus vertical component of drag, as illustrated in the answer."  -- no,, in a steady-state climb, net vertical aerodynamic force, the vector sum of all upward and downward aerodynamic components, must be equal in magnitude to weight (and opposite in direction, of course.)

This question is purely a definition issue, and the answer is 'yes' or 'no' based only on which definitions you use. In Newtonian physics, a lot of complex interactions are modelled as single, lumped vectors which we call "forces". These forces share nice properties with vectors: notably, that we can decompose vectors into multiple vectors, or sum multiple vectors into a single vector. One of the main reasons to do so is to decompose a vector into components parallel to some (typically orthogonal) coordinate system.
An important observation is that there is no 'true' way of representing the forces acting on the airframe. While some decompositions are more popular than other, all are equally valid (if done correctly). I will take two examples, one of which arrives at your conclusion 'yes', the other 'no'.
Example 1. Decompose the aerodynamic forces on the airplane, parallel and orthogonal to the flight path. Call one 'lift', call the other one 'drag'. Let's assume 'thrust' is also parallel to the flightpath. Weight is represented as a single vector, orthogonal to Earth, and is not decomposed along the flight path. Now take all the forces that we decomposed along the flightpath, and again decompose them but now orthogonal to Earth. Now, only look at the forces pointing 'up', which in a climb (but not in a descent) removes the vertical component of the  'drag' vector, and compare it to the weight vector. With this elaborate procedure, we can conclude the answer is 'yes'.
Example 2. Combine all aerodynamic forces on the plane in a single vector instead of decomposing them into lift and drag, called the 'net aerodynamic force'. Leave the thrust and weight vector unchanged. Again, we decompose all vectors along the Earth reference frame. Now, we find the sum of all the upward components is exactly the same as weight. We can conclude the answer is 'no'.

Note: the net aerodynamic force is shown in the left diagram for illustration only to show that it is the sum of lift and drag, and is not actually part of the force balance for example 1.

Would this case be example 2 as well?

@Koyovis In that case there is no lift, only drag, so there's no difference in the net aerodynamic force and the decomposed lift and drag vectors. If your climb is steep enough, example 2 changes from 'no' to 'yes'. (1/2)

Either way, this highlights why I wonder what the practical value is of your question at all, because the answer changes according to definitions and flight conditions. What exactly inspired this question? (2/2)

The downwards pointing drag is there as well in shallow climbs, if small enough like with GA planes it can be neglected. When working with aerodynamics of helicopters & military jets, it is very apparent that this force needs to be considered in questions about lift & climb.

@ymb1 No idea either. I lumped parasitic drag with the rest of aerodynamic forces, because the point here is to show that the definitions you use to separate the forces acting on the airframe determine the final answer, not to elaborate a complete description of all forces acting on the aircraft.

This is the right answer, for reasons @Sanchises explains well. It all depends on the definitions. If we are very pedantic about the question, we can see that in a steady-state purely vertical climb that the net thrust is weight PLUS drag. So that's a way to say "yes". But if we get away from the "gotcha" nature of that nuance, then we have a simple Newtonian mechanics problem which stipulates that all forces must balance in steady-state motion.

@Robert Look at the vector diagrams. They're too scale. Excess thrust is available. Forces balance. What are you saying?

What I am saying is forces balance after upward velocity reaches steady state in all cases.  No need for two diagrams.  Therefor total upwards force equals weight plus vertical drag.  No need to separate rocket case, aircraft case, helicopter case, case 1, case 2...  Amazingly, the answer is a simple modification of the 4 forces they teach in flight school. Even more so, how much more thrust is required to offset tilted vector. (I suspect this may be the real secret to "laminar flow" wings, tilting the lift vector forwards a bit).

@RobertDiGiovanni The two diagrams show the exact same case, with the exact same forces, but just represented slightly differently to show that which definitions you use changes the result. Please make an honest effort trying to understand what I'm writing instead of jumping to conclusions.

I'll buy that.  Your diagrams (and Quiet flyers) inspired mine.

Yes.

Vertical forces are forces in the earth axes reference frame. So remaining in this reference frame, assuming $\gamma$ ≈ $\alpha$ in a steady climb, the vertical forces are:
$$W + D \cdot \text{sin }\gamma - L \cdot \text{cos } \gamma - T \cdot \text{sin } \gamma = 0$$

Pointing downwards = + = $W + D \cdot \text{sin }\gamma$
Pointing upwards = - = ($L \cdot \text{cos } \gamma + T \cdot \text{sin } \gamma$)

So in order to continue climbing, the total upwards vertical force - consisting of a combination of Thrust and Lift - must be larger than weight by a factor of D * sin $\gamma$
Notes:

Upwards aerodynamic force is often called lift. But lift is defined in the airframe aerodynamic axis, and tilts with the direction of the airspeed. So for a fixed wing aeroplane in a steady climb, total vertical force is higher than weight, but lift is smaller than weight.
If the upwards force changes, the climb speed changes accordingly. There is an acceleration, causing a change in aerodynamic drag which stops when the forces are in equilibrium again.

Only in a constant density atmosphere. In reality the airplane loses climb speed with increasing altitude, so there is a vertical deceleration term which reduces the vertical force ever so slightly.

Lift is NOT total Upwards aerodynamic force! It is the component of all aerodynamic forces perpendicular to the plane formed by the wings and the flight path of the aircraft. If in a climb, then the flight path is inclined upwards and not vertical, so lift (if it to have any same meaning) is also not vertical. You CAN define it any way you want I suppose, but defining it as the vertical component when the aircraft is in a steep climb or descent, (or for that matter a steep bank), serves only one purpose, to confuse.

@PeterKämpf The plane is also burning fuel, reducing the weight over time as well. The plane in question is climbing at a steady rate.

@CharlesBretana Thank you for pointing out the value of asking this question twice, once about lift and once about net upwards force. Mentioned in note 1. in the question.

And what is net upwards force in an F-16 in a pure vertical climb? Do we count the force of the dynamic air pressure on the front surfaces of the aircraft as negative upwards force? what about the force  of air pressure on the rear surfaces of the airframe? They are of course pushing the aircraft upwards. Are they "Lift"?  Breaking up aerodynamic forces into defined components is done to create understanding and simplify and enable calculations. Applying these arbitrary definitions in scenarios where they do not help to do that only confuses.

@CharlesBretana It’s a matter of axes definition. Above question is referenced in earth axes, avoiding the confusing axes tilt of the aerodynamic reference frame.

I Guess you could say that. What I'm saying is that how you define terms (what axes you use to define them) not only determines your analysis, but sets the context for the scenario your analysis will be meaningful or useful in. Defining terms inappropriately, (using the "wrong" axes), results in meaningless or useless conclusions. In this case, what is significant is that vectoral acceleration MUST be directly proportional to the vector sum of ALL forces (F=ma)! Any definition of forces or choice of axes that does not clearly show that is pointless or wrong.

@CharlesBretana In 3 dimensions,  vectorial sums are constructed in each of the 3 axes. The axes are perpedicular to each other. The pic in the question shows the relevant forces in earth axes z-direction, the point being that there are three vertical forces, not the two that are usually considered. Note that it depicts the situation at constant velocity airspeed and - climb speed.

@Koyovis, Actually, vectoral sums can be calculated in any basis set. Cartesian axes are just one of the way to do so. forces can be broken up into components based on polar or spherical axes as well. They don't even really have to be normal to one another (although non-normal choices of axis introduce significant complexities). Also, when all relevant forces lie in a 2-D plane, it is only necessary to perform the calculations based on a choice of two axes that also lie in that plane. There is no need to include a third axes when there are no forces in that dimension.

@Koyovis,  Finally, the choice of which axes to use when breaking up the forces into vectoral components is critical. Choosing a set of axes that increases the complexity of the necessary calculations, or obfuscates the underlying physics, just because "Those are the standard axes" or "That's the way we always do it", is probably not the optimum way to do engineering. The choice of axes should be based on the problem to be addressed.

If we take the op's question, as worded, (specifically his use of the word Vertical, then in fact there are actually only TWO dimensions in play here, the vertical axis, (as you say, the z-axis perpendicular to the earth's surface), and a horizontal axis parallel to the aircraft flight path. And there are two forces that need to be broken up into components along those axes. The force imparted by the engine thrust, and the total aerodynamic force on the airframe. The vertical components of these two forces must indeed be equal and opposite to the fictitious "force" of gravity.

..plus the vertical component of the airflow resistance.

Well, yes, But I would include "airflow resistance" as just a part of the "Total Aerodynamic force" on the airframe. We tend to break it up in a myriad of ways, "Lift" vs "Drag", then we break up "Drag" into "Form drag" and "Induced drag", where induced drag is component of total lift that is parallel to the flight path, etc. etc. But in actuality, it's all just the result of air molecules hitting the surface of the airframe.  Each molecule that bounces off the airframe imparts a force normal to the surface at that point, proportional to the molecules change in momentum.

The total aerodynamic force is just a vectoral sum of all these individual molecular forces, over the entire surface. Then, depending on what problem we are trying to solve, or situation we are trying to understand, we break up this vector sum into multiple components, oriented in directions that make sense for that specific problem or situation.

Yes, always, unless aerodynamic drag does not exist.
For winged aircraft, the above is impossible, therefor the answer is yes, always.
It is important to realize excess thrust is required to climb.  Excess thrust closes the weight/lift triangle but does not account for aerodynamic drag, which equals the amount of additional thrust required to maintain airspeed.
Add this "handle" (in the direction of flight) to the closed vector diagram for vertical lift and voila!  There is your vertical aerodynamic drag component (decomposed from the aerodynamic drag vector).
Always there, in any climb, for any aircraft.
An aircraft that has a thrust to weight ratio of less than one simply must use a ramp while maintaining airspeed against drag.
The combination of excess thrust and lift support the weight, enabling steady state flight with 0 acceleration (from Gravity), while the remaining thrust at a given velocity opposes aerodynamic drag (part of which is being used to create Lift).

Like this.
In level flight only around 300 lbs of thrust is needed, as the far more efficient wing can now bear all the weight.

If the net forces are zero, the movement will be steady, as per Newton's second law.
If upwards vertical forces equals weight, we will have net zero vertical forces and a no vertical movement (hover).
If the total upward forces are greater than weight we will have a vertical acceleration until drag brings velocity to a steady state.
If all vertical forces = weight the aircraft may be rising, hovering, or descending with 0 acceleration.

Thanks Robert for the edits! Could you check again to see if there is a possible conflict between paragraph two and four. If so, I think it is pragraph four that should be kept.

In paragraph 2 "downward" force from drag = 0, so 2 and 4 do not conflict.  Please excuse my passion (in editing).  Paragraph 4 covers all 0 vertical acceleration cases, indeed, vertical forces essentially make the plane "weightless" leaving the drag from velocity.  (Now we can think about airships too).

Ah, I see, drag has a downward component in steady climb. And we presume it is a climb. And we exclude downward compents from upwards vertical forces. Then its the slowest climb = hower. Couldn't it still give the reader the expression that zero force always have to give a standstill and not as in Newtons second a steady movement?

An object in an environment with drag and 0 net force will slow to 0 velocity. (Unless drag force is included!).  Since we are comparing all upward force to weight, if there is upward velocity, there is a downward drag component, therefor...


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How can an altimetry system error during cruise (500 ft at 35000 ft high) cause an insufficient fuel situation?
So my boss gave me a question today as my daily quiz: Let's say an A321 is cruising on a 6-hour flight. It reaches FL350 30 minutes after take-off. After another 8 minutes there's a discrepancy between 2 altimeters. PFD 1 reads 35000 ft while PFD 2 reads 34500 ft. The pilot decides to make an air-turn back due to insufficient fuel, despite having enough fuel filled while on ground.
The question is, did the pilot make the right decision?
I've done some research. The difference between 2 altimeters is not supposed to exceed 245 ft at 35000 ft. But I don't know how that is related to fuel shortage. Worst case scenario the static port is contaminated.

Hello Light Le, welcome to Aviation.SE. Nice quiz question, if your boss has more of those, keep us informed :-)

@DeltaLima sure I will. My boss loves to grill me on these kinds of trivia and won't leave me in peace until he gets a satisfactory answer (his direct quote).

If the aircraft is cruising at FL350 it is operating in Reduced Vertical Separation Minima (RVSM) airspace (between FL290 and FL410).
With an altimetry discrepancy between the two altimeters of more than 245ft, the aircraft is no longer RVSM approved. It shall therefore descent or climb out of the RVSM altitude block.
For the Airbus A321 the operating ceiling is below FL410, so the only option is to descent to FL280 or below. Since the fuel consumption in those altitudes is significantly higher the aircraft is unlikely to complete its 6 hour flight with the amount of fuel on board. Therefore the decision to return is likely the right one.

Awesome :-) QED

Excellent :) I'm not an aviator, but this style of question is great for technical and procedural testing in interviews

What can happen if he just keeps flying in that area?  Would he get arrested when he lands?  Or potentially crash his plane?

@Nelson make it as a separate site question. "What happens if a pilot flies in RVSM airspace with a craft not RVSM approved?"

@Nelson - If the discrepancy is known outside the cockpit, the pilot could be grounded for this kind of infraction in a commercial airliner. RVSM essentially means 1000 ft vertical separation between aircraft. If your altimeter is off by 250 feet, and the other guy's altimeter is off by 250 feet the other way, now you're passing within 500 ft of another plane even if you're keeping the needle right on your flight level; that is generally the absolute minimum safety margin for flight. Deviate even a little further and now it's a "near miss".

If he's flying in RVSM space outside the limits of altimeter discrepancies, then facing administrative action is literally the BEST possible outcome.  A head-on collision with another plane, resulting in mass casualties is an entirely realistic possible outcome.  He won't worry about getting arrested after that.

Mid-air collision like this: https://en.wikipedia.org/wiki/Gol_Transportes_A%C3%A9reos_Flight_1907

@Nelson you should ask your question as new one (as suggested by Mindwin)

Why would they have to make the A321 return to base, instead of simply clearing a block of airspace surrounding it?

@Sean because that is not simple at all to do unannounced on a 6 hour flight.


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Why doesn’t lightning punch a hole in the fuselage?
Electric arcing creates a huge temperature in the air around it. When lightning strikes planes, why doesn’t this high temperature air melt/deform the aircraft skin? Also can these high temperatures ignite fuel in the wings?

What is arcing?  I mean, what is arcing between?

Potentially, yes, but this would be a very rare event. The principal reason that lightning does not create damage to an aircraft is that the airframe, particularly aluminum airframes, are electrically conductive, and carry the charge easily around them with very little resistance, which would otherwise create heat. Essentially, the airframe behaves like a Faraday cage and protects the contents within from damage. The airframe is also designed to dissipate static electric charges through static wicks, lightning runners and lightning spikes on the exterior of the skin.  Fiberglass parts such as radomes are protected by lightning runners which carry the charge and protect electronic components underneath.  Composite aircraft made from e-glass are constructed with a stainless steel mesh ply imbedded in the skin which channels and carries electric charges.
Lightning strike protection is a critical part of aircraft certification under parts 23, 25, 27 and 29.  Very rarely has lightning caused any kind of an accident, and commercial aircraft are frequently struck by lightning every year and suffer almost no ill effects from it.

The question is about "high temperature air" so I do not see how the conductivity of the airframe answers that part of the question.

Those kinds of air temperatures last in the timeframe of milliseconds. Not exactly enough time to heat up and melt and/or deform the skin in.

Just because the temperature is high, doesn't mean that it is magic and blows up like a bomb. The spark from a grinder is melted metal, but the total heat energy is so low that the workers usually don't need any extra protection around it, even if it is sprayed all over their body.

Re Nelson's comment on grinder sparks, I can show him my glove where the spray burned a channel through the back within minutes.

"Composite aircraft made from e-glass are constructed with a stainless steel mesh ply imbedded in the skin which channels and carries electric charges." I've wondered about re-certification as they age, embedded metal mesh with 20 years of occasional desert heat sounds like a recipe for gaps in protection.

@AndyDent sure, but that it took minutes to burn that channel is supporting evidence for the claim that microsecond exposures aren't going to matter.

@AndyDent I'm sure if you dipped the glove in molten steel it'll combust much faster.

Side note: commercial aircraft are frequently struck by lightning every year and suffer almost no ill effects from it. Granted, but what about the passengers? See: Frequent Flyers Could Take a Hit of Radiation from Lightning and 'Dark lightning': The mysterious radiation that might be zapping air passengers mid-flight An issue [or not]?

@CraigEstey Refer to the Radiation Dosage chart by XKCD.

Though there must have been some lessons learned in the early days of aviation...

Don't forget that the aircraft is moving through the air at speed. The piece of air that was superheated is far behind the aircraft very quickly!

The other answers mention aircraft have a high electrical conductivity.  However, air has a much lower electrical conductivity and does get hot, as you mentioned.
The hot air causes little damage because:

The electrical current only heats a small area of the air for a brief period.
The air expands quickly, causing it to cool quickly.
The aircraft moves quickly, so it is only exposed to the heat briefly.  After that, it is exposed to very cold air that can remove the heat.
Since the aircraft skin is electrically conductive, it also conducts heat away from the hot area.

Also the metal has much higher heat capacity, so the same amount of energy heats it less then it heats the air, and good thermal conductivity, so the heat spreads quickly further reducing the thermal stress on the point of contact.

Bullet 1: Actually, part of the point is that a lightning strike is much LESS concentrated, and much shorter than pretty much any other arc, whether that be an accidental wiring issue, or from some sort of welding.
Bullet 4: Electrical conductivity and temperature conductivity are not the same thing.

@MikeB while not being the same thing, good electric conductors are usually also good thermal conductors (see an answer at physlink.com, or the sortable list on Wikipedia) it's not working the other way.

Why doesn't it punch holes in the fuselage? It can.
Can it ignite fuel in wings? Yes, lightning creates sparks that can potentially ignite fuel vapors. The liquid fuel itself does not ignite, but the vapors can.
Both of these things are extraordinarily rare, however. Planes are specially designed to be resistant to lighting strikes, and most lightning strikes result in little to no damage.

All true. But please speak to how rare this is. I have to fly with people who read this.

Each airplane is struck, on average, 1-2 times per year:  https://www.weather.gov/safety/lightning-planes   https://flightsafety.org/wp-content/uploads/2016/11/asw_jun10_p18-23.pdf

It should be noted that damage like shown in the linked picture is, at the moment it happens, completely inconsequential. The structural elements around it are strong enough to prevent the hole from growing bigger and the outflow valves are bigger than that so they'll just close a bit and the pressurization can keep up just fine. Only long term the damage would spread mostly due to metal fatigue, so it has to be repaired after landing. But it wouldn't cause any immediate danger.

@JanHudec That hole looks big enough to put my hand in.  I'm not understanding how pressurization wouldn't cause a lot of extra work as now air will be blasting out of that hole which will need to be pumped back in...

@Michael The cabin is not sealed to keep air inside, but the engines or compressors are continuously pushing quite a large volume of air into the cabin while it leaks through all the imperfect seals and the outflow valves. Quick search says the outflow valves on a 777 have (the two of them together) area about 90 in²/580 cm². You can definitely put your hand through that.

@JanHudec Ah that makes sense, thanks.  Incidentally, if this "engines or compressors pushing a large volume of air into the cabin" the reason why at times the air has a bit of a distinct - I was to say almost oily - smell to it?

@Michael The hole depicted is in the skin of the wing, not through the pressure hull. The insides of the wing are not pressurized.

@Michael, Yes, most jet aircraft use ‘bleed air’ taken from the engine's compressor stage for pressurization and with that a bit of engine oil can get in. It's part of the reason why Boeing returned to using separate compressor in 787.

@Chris you sure that's a wing? It looks far too rounded to be a wing to me. But, I've been wrong before...

@FreeMan It looked like a wing to me, but now that you mention it, that looks like the tail in the background, which would make it the fuselage after all.

@Chris I thought that was the tail in the background, too, but it looks too long and has too many fairings along the bottom for me to want to claim that it's the horizontal stab. Again, though, I could be wrong. ;)

When electric current flows through a medium, the amount of power dissipated is inversely proportional to the resistance of the medium. This is why the heating element in a space heater heats up, but the power cord leading to it doesn't (much): the element has a much higher resistance than the cord. This is also why using too small of an extension cord for the load can cause a fire: thinner conductors have a higher internal resistance.
Air has a much, much higher resistance than the aluminum, composites, etc. that they use to build planes, and hence gets much hotter during a lightning strike. This is easily seen in videos of planes being hit by lightning, where the air glows from the heat and the plane doesn't. Furthermore, heat can't transfer from the air to the plane because it's almost immediately blown away by the slipstream.
Of course, that's not to say that it's impossible for lightning to cause damage, because it absolutely can. If the current finds a spot of higher-than-average resistance, or gets concentrated to a point, it can (and has) caused significant damage.

Wait, composite materials like carbon fiber have low electrical resistance?  Apparently that's true: https://carbonfibergear.com/blogs/carbonfiber/is-carbon-fiber-conductive#ol4   - neat.  Unlike fiberglass, which being made of glass is an insulator (one which doesn't ionize nearly as easily as air).

Fun fact: the amount of current that the power cord of a space heater carries is equivalent to about one typical lightning bolt every second. Passing that amount of current through air produces tremendous amounts of light, heat, and sound; passing that amount of current through a copper wire does practically nothing.

@TannerSwett I don't know where you picked up that little factoid, but it's not exactly accurate. A typical lightning bolt is well over 30,000 amps, while a typical household electrical socket is limited to 15 amps.

@HiddenWindshield Yes, now that I think about it, my comment is really not accurate at all. What is true is that the amount of charge carried by the power cord in a minute (about 700 coulombs) is a reasonable amount of charge for, say, 40 lightning bolts. When I wrote that comment, I was thinking that the actual instantaneous current doesn't matter—in other words, I was thinking that 30,000 amps for half a millisecond would be practically the same as 15 amps for one second—but now that I think about it some more, that's not true at all. Sorry for the mistake!

Wire heating from intermittent loads will be roughly proportional to duty cycle, but the square of the current, so reducing the duty cycle by a factor of 100 would only allow a 10-fond increase in current (rather than 100-fold) to keep the same level of heating.

@TannerSwett you're right but you're confusing people. The reason the power cord would do "practically nothing" like you say is because the lightning has to do much less work to travel over the copper than over the highly resistant air. It's why lightning rods work. Which is what you're building with this thought experiment. Extension cords don't typically get bigger than 10 gauge (AWG). Lightning rod wire must be 6 gauge (bigger). Typical extension cord has 3 wires. 3 10 gauge wires gives 5 gauge. So yeah, if you get lucky and the lightning hits all three wires evenly, "practically nothing".

"Air has a much, much higher resistance than the aluminum", lightning in air doesn't flow in air as a gaz, but in air plasma which is a very conductive.

@mins Plasma isn't "very conductive". Sure, it's a lot more conductive than non-ionized air, but it's still much less conductive than aluminum. Also, the current is what creates the plasma in the first place, so there must be at least some current flow before the plasma can exist.

Air ~$10^{-12}$ S/m,  carbon fiber ~$10^{-2}$, lightning plasma ~$10^{4}$, Al ~$10^{7}$. There is not much current before plasma appears, insulating molecules are broken by voltage, and turn into plasma quickly as current starts increasing and becomes a precursor.

@mins So, aluminum is about one thousand times more conductive than lightning plasma? Sure, sounds about right. Also, conductivity is only one factor in determining resistance. Resistance is also inversely proportional to the area of the conductor. A lightning bolt is only about an inch (2-3 cm) across. Planes are bigger than that, so the actual resistance is much lower.

Yes aluminum is more conductive than plasma, but the point is plasma is one million times more conductive than carbon fiber composites used in modern planes. In addition carbon fiber materials have a low thermal conductivity slowing down heat dispersion. Currents observed during lightning strikes on actual aircraft were up to 24kA, but for very short times. I think the reason is the short time rather than the large section of the conductor. Two studies: 1, 2.


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Why do airplane indicators start at 12 (o'clock), unlike cars that start at 7?
A car's speed and RPM indicators usually start at 7, lower left, but airplanes' altitude and airspeed indicators start at noon. What's the origin of this convention?

Aircraft RPM meters typically start the same place the automotive ones do. VSI typically points either at 9 o'clock or 3 o'clock. Many gauges start at 10 o'clock to 2 o'clock (temperature, oil pressure)...

In (motor)gliders you'll often find the airspeed indicator resting at 6

Aircraft indicators don't like to get out of bed till the crack of noon.

Most old-style indicators (RPM, cyl. head temperature, outside air temperature, oil pressure, etc.) are geared in such a way as to present their full operating range over a needle swing of 270 degrees. It is also common for the gauge mechanism to be oriented in such a manner as to put the nominal operating point somewhere near the middle of the needle's swing arc, where the gauge mechanism introduces the fewest errors in the reading.
Early (crude!) altimeters were also 270 degree swing gauges (i.e., barometers), but this made it impossible for the pilot to read that gauge to an accuracy better than perhaps 500 feet- which is insufficient for modern use. Later versions of the altimeter had gear trains inside them to magnify the needle movement enough to read the gauge to an accuracy of ~50 feet, but this in turn dictated that the gauge range cover multiple revolutions of the needle around the face of the instrument... like a clock.
Since we are all used to reading clocks that start at the "12:00" position, the multi-turn altimeters were set to start at 12:00 too and geared so one full needle revolution corresponded to one thousand feet on the "minute" needle, and ten thousand feet on the "hour" needle. This worked fine for small planes that rarely flew higher than 10,000 feet.
Registering multiples of 10,000 could be done with a third needle but a three-needle gauge isn't the easiest thing in the world to quickly scan. A better solution is to have a window cut into the gauge face behind which altitudes greater than 10,000 feet are displayed as numbers instead.
As an aside, in older American-made car speedos (1960's-era), 60 mph was most commonly positioned at the 50% scale point ("straight up" on the dial face) to give the driver a quick visual indication of 60, without having to read the numbers. When the US federal maximum speed limit was reduced to 55 during the oil crisis of the mid-1970's, 55 was then positioned at the straight up spot.

Also, older automotive speedometers weren't circular; the needle traversed from left to right, roughly from what would be 11 o'clock to 1 o'clock, with neither the top arc nor the bottom 2/3 of such a dial being present.  When you consider the height of automotive instrument panels, that flattened display makes sense -- less real estate to work with.  As that wraps into a circular gauge (ask the automotive historians why cars went to round dials), the zero becomes left and low (7 to 8-o'clock ish), and the max becomes right and approaching horizontal (2 to 3 o'clock ish).

There is a rationale to scale and orient "health" gauges (temperatures, pressures, RPMs, etc.) such that the pointers all align in the same direction, such as 3 o'clock, when values are in their "green" or nominal range. This can help make an off-nominal condition stand out more in a quick visual scan. It can put zero and maximum values at seemingly odd positions on the dial, but that is functionally unimportant.

"Federal" speed limit refers to which country?  Quite a few countries have federated government, but you didn't say which one you're talking about.  (Admittedly, not many places use mph, but even so...)

+1, but when I was flying gliders in a metric country most altimeters had zero at 6 o'clock.

As for what could be done, most "modern" analogue readouts are now assymetric, the first 90 deg covers less range than the next 90 deg. Or for the barometer types they are usually separated into "OK" range and "NOT OK" - which each can have a separate sensitivity. That is at least top of the line where I work, for the odd and analogue things where the computer doesn't take care of it all.

@StianYttervik, god dag, og where do you work? -Niels Jakob

A little off topic on cars: it actually wasn't 60 MPH at 12 o'clock but if you looked carefully it was 100 km/h (62 MPH). It was largely seen in Japanese vehicles because Japan didn't have any speed limit over 100 km/h until 2017. It's not for the US because a 60 MPH limit is relatively rare, limits typically end in 5s: 45, 55, 65, or 70.

@TobySpeight actually, most countries which are federations do apply a nationwide speed limit. For example, in Russia there is a national speed limit of 90 km/h outside city limits (110 km/h on highways). (Although, Russian traffic code has no fine for speeding by 20 km/h or less, so many motorists interpret it as speed limit of 110/130 respectively.)

@RalphJ  "Also, older automotive speedometers weren't circular"  Say, that's not really correct.  The "flat style" of speedo was popular at a certain time (around the "massive tail fins" era of cars!) but all early car speedometers were simply circular dials. (You can easily google "history of car speedometers", it's a huge area due to the interest in early 1900s antique vehicles.)

To put it another way, imagine that your car's speedometer went to 40,000mph - try holding a steady 60 when the nearest gauge ticks are 500mph apart.

Amazing explanations. That’s what i was looking to hear thanks everybody !!!


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Why does the airplane not 'feel' the crosswind?
Apologies for multiple pictures but I thought they would depict my question more clearly than a lengthy description.
Image 1: Why does the aircraft still follow the straight path even if the wind is blowing from left to right?
Image 2: Why does it appear that the aircraft is sliding?
Image 3: Why is there no need to apply any rudder in the presence of a cross wind?

The straight line the aircraft is following is the railway line marked in the centre of the map running North-South.  Take a look again at the aircrafts position moving from South to North.  There is no East-West movement.

The aircraft doesn't "feel" the wind for the same reason you don't feel moving 1400 miles per hour as the earth rotates - the plane is in the wind/air so from the point of view of the plane there is no wind just like you are on the ground so from your point of view the ground isn't moving 1400 miles per hour.

Consider the north is pointing to the figure top. At the airplane flying level the wind can be blowing to east and at he clouds level it's blowing to south that's is one of motives clouds are not good for reference points. For the sake of the question consider at plane level the air is not moving but the earth below it is moving and you ill find your answer.

No apologies needed for the pictures: They nicely illustrate the effects of flying in a crosswind. The aircraft is pointing into the wind direction, the clouds nicely illustrate the left-to-right wind and the ground track is bottom-up; again easily to read from the sequence of pictures. And whoever made them easy could have picked an uglier aircraft that the Spitfire for the illustrations.

How would a pilot land an aircraft when there is heavy downwind. Reducing the speed might stall the plane. So again, he does nothing to counter the heavy downwind while landing?

Those pages look they are from the 1930's or 1940's.  Seriously, we're supposed to turn rapidly between the pages to get a cinematographic effect?  It's all different now, we have GPS, so none of this matters any more.

When dealing with crosswinds and drift, I find the easiest way to understand the concept is to imagine a fast flowing river of water. If you tried to swim straight across a fast river you will end up downstream. You don't "feel" the current. You just keep swimming and the current carries you.
Any time you have wind it is like a river of air flowing over the earth.
Traveling through a river of air is exactly like traveling through a river of water.  You don't go where you are pointed, you go where the current carries you, plus the direction you are swimming.

So, can I say that headwind makes the aircraft go slower while tailwind makes it go faster? (I thought its the other way round, since when you have headwind, you have more air flowing over the surface which makes you go faster)

Yes, you can say that. The other way speculation is incorrect. Tje aircraft maintains certain indicated airspeed and true airspeed, but due to the wind drift the ground speed is different just because the plane is moved passively by the wind.

@user2927392 your problem is that you are mixing up airspeed with groundspeed. They are related, but I suggest you look up how.

@user2927392 Assuming a constant airspeed, then yes, you can say that about the groundspeed. Also, just in case you aren't familiar with the terms: airspeed = speed with respect to the air around the airplane, groundspeed = speed with respect to the ground below the airplane.

so, is the same motion of air (wind) the cause for turbulence? Can I say turbulence is like an odd bump in the road caused by headwind (or tailwind?)

Turbulence is not headwinds or tailwinds.Turbulence is caused by abrupt changes in direction and speed of the wind, by passing over obstacles like mountains, being heated by the ground from below, and also by clouds and cloud formation .

The main reason is that you are, in fact, flying straight through the air, and the air is moving sideways. So the plane isn't really being pushed sideways or anything, it's just flying straight through something that is moving...
As a simple example, if you are flying north at 150knots (the red line below) and the wind is blowing east at 20 knots (the blue line), you are going to end up flying just every so slightly North East at about 151knots (roughly). But, to your plane, you're just traveling north at 150 knots, the wind is adding the last 1 knots.

The bottom line is that your plane flies in a straight line through the air mass, and the air mass moves as well.  You have to add these two velocities/directions together in order to find your true velocity/direction relative to the ground.  This is normally called a "vector" in the aviation world, and if you do a bit of study into those, I think you'll have a much easier time understanding those charts.

A change of reference frame may help to understand. So you sit in the plane and fly as usual, the only effect of wind is drifting the ground in the direction where it comes from. Headwind moves the ground forward, tailwind backwards. Sidewind to the side, so you compensate not by applying continuous rudder, but just by pointing the nose slightly to the side.

As for your question about why no rudder is required, there are two methods for handling a crosswind.  One is "crabbing" and the other is "slipping."
When you crab, you point the nose of the airplane into the wind as described in the other answers and in your pictures.  This is an efficient way to deal with a crosswind over long distances, such as when cruising.
During landing, if you are still crabbing when the wheels touch down, it's very hard on the airplane and is called "side-loading" the landing gear.  Instead, a slip is used.  In a slip the ailerons are held towards the wind and the rudder is held away from the wind just enough to keep the nose from turning.  In this way the longitudinal axis of the airplane is still kept in line with your flight path, but some of your lift is being used to "pull" you up-wind and thus counteract the crosswind.  On most airplanes this is the method used right before touchdown.
On the final/short final phase, pilots have different preferences for crab or slip technique.

In general: as Jay Carr visualized with his interactive graphic, velocities (speeds) are vectors (this is rather physics than aviatic, but if pilots consider physics they tend to live longer). Vectores have a value and a direction. So you can visualizes them by an arrow (lenght visualizes the velocity). Contrary to e.g. temperature which is a scalar measure, it has only a value but no direction.
So to add speeds you must consider values and directions.
Now to your questions ...
ad Image 1)
The wind is blowing from the left. If the aircraft would aim to the north ("heading north"), the cross wind would move it to the right, so it would pass the city more to its right (if map is oriented "north up" than it would pass the city east).
BUT, the pilot is smart: he compensates for the crosswind by pointing the aircraft's nose slightly "into the wind" (look, where the aircraft's noise is pointing to). Durign his flight planning he might have calculated the "offset" comparing desired track and forecasted wind. Once in the air he could also find out by applying a heading into the wind and check whether the aircraft moves along the desired track.
ad image 2)
the plane is sliding (well, it does not touch the ground) because its air speed (caused by the propulsion of the prop and being straight - visualized in the image by the wake vortices and its wing tips ) points to its noise (the prop is mounted without pivot), plus (plus in a sense of vector addition) the wind from the left result here in a track following the railways.
ad image 3)
the wind from the left moves the whole air which surrounds the aircraft (a static wind, no gusts). So the while the aircraft moves through this air mass, the air mass itself (and with it the aircraft) moves along the wind direction. So you do not need rudder to hold that direction (heading). You only have to apply rudder to change the direction. So the pilot did apply rudder for some seconds when he noted that he needs to compensate for the crosswind, just until the aircraft changed its direction from north to slighly north-north-west.
The author says "the aircraft does note feel the crosswind" - yeah, due to its static effect you could not tell without any reference (to the ground, to your GPS or whatever navigation aid) that there is a wind at all.
To bring Wurzel's answer into the context: yes, if you want to land then you could use the same maneuver as shown in the 3 images (called crab, because this animal walks sideways), - or if you could fly straight along the runway direction but compensate for the crosswind by lowering the windward wing.
Hope it helps
Holger

authors: Vladimir F Героям слава


text:
Does a strobe light shield act like a wingtip fence [or better sort of vortex generator]?
I was told it is a shield to avoid to blind the pilot during night flights when the strobes are on. Now, knowing the primary reason is, as far as I know, the above mentioned:

Does anyone know any official document where the definition of this device is mentioned?
Does anyone know any reference where the effect on the wing tip aerodynamic is studied? Considering different speed regimes?
Does anyone know if there are any regulations on this small device?

As far as I know, this device can be founds on the following aircraft:

King Air
T-6 Texan
Embraer Legacy 600

As can be seen, it can be found on models with and without wing tip devices.
One thing that should be mentioned on aircraft fitted with winglet (not wing fence at the tip as those found on the A380), is that during the pull up manoeuvre, at take off, a mild/severe separation is generally observed on the inboard part of the winglet. Could the strobe shield fitted in the Embraer Legacy 600 mitigates this by re-energising  the boundary layer. It would be easy considering the amount of cross flow at the tip.
Any official paper or informal discussion anywhere in the web would be very useful.
More pictures can be found on these links:
http://79ft.net/tag/mechanicing.html

I think in some conditions it can be that it triggers the separation and it creates a vortex, but has it been designed to do so?
I don't think so, while I imagine it would require intensive testing in low speed/ wind tunnel to verify the design, and I am not sure it is "worth", but maybe that part of the wing is more "critical" for stall recovery than I think...

@GHB Typically you want the wing to stall from the root and progress towards the tip, so that the ailerons stay effective for roll control. I'm guessing that once you get past the aileron surfaces, the wing being stalled or not is less of a factor in controllability, and I'd venture to say that having a vortex generator on the very tip is going to do little for performance.

@Ron Beyer, good arguments, so a vortex triggered at the tip could, potentially, by shifting inboard, help reattaching the flow over the outboard part of the aileron.
I am not sure quantitatively the effect of this reattachment be and, most important how "robust" that would be, to be considered a mechanism triggered by "design" (so sought by engineers, tested and validated).
For the same argument of "robust design" I am not sure what is the "value" to such a design from a performance point of view.

Based on the comments posted by the others, which more or less agreed with my thinking and confirmed other hypothesis I have made, I would appreciate if anyone would post any official papers/documents to support their arguments.
Here the difficult thing I have found is to find 'official' statements.
Is it possible this is something for which no regulation at all can be found?

As is said in an answer below, this is not a vortex generator-like device nor will it function as such. It acts more as a winglet.

Vortex generators are designed to trip/refresh the boundary layer. A turbulent boundary layer delays massive separation more than what laminar boundary layer can do. As such it could work as VG unless somebody find a good argument which proves my argument wrong. It is my opinion that this device most definitely will NOT work as a winglet as you (SMSvonderTann) are suggesting. My question, I repeat here, does anyone know any official studies on the topic? Again, this device has the size of a vortex generator (or very small fence), whether it was design as such or not, we still do not know.

To be more precise, if we want to see a wing fence (now we are really playing with the definition) acting as a sort of winglet, as suggested before, we have to look at these two pictures http://john2031.com/piper/pa-16_clipper/alaska/n5876h_4.JPG and http://www.flysquirrel.net/sterling/sterl08.JPG. But in these cases dimensions are really different. Hope this helps.

It is a long structured question, I will try to keep up with the answer. As first I would like to cite some sources.
I compared both CS23 and CS25 looking for the certification requirements on the lights. They are almost identical. If you want to check them yourself the keyword you want to use is "Lights". The paragraphs of main interest are from 1285, to 1401, three pages in total. That said.

I was told it is a shield to avoid to blind the pilot during night flights when the strobes are on.

The light on the wingtip in this case are both position lights and anti-collision lights, the strobing light you refer to.
The Position lights: are described in paragraph from 1285 to 1391. There are requirements on the shielding of them and the intensity of the lights from different radians as well as the colors of them, but none from an aerodynamic point of view.
From a design point of view the shield should define some dihedral angles:

(b) Dihedral angle L (left) is formed by two
  intersecting vertical planes, the first parallel to the
  longitudinal axis of the aeroplane, and the other at110º to the left of the first, as viewed when looking
  forward along the longitudinal axis.

(c) Dihedral angle R (right) is formed by two
  intersecting vertical planes, the first parallel to the
  longitudinal axis of the aeroplane, and the other at
  110º to the right of the first, as viewed when looking
  forward along the longitudinal axis.

(d) Dihedral angle A (aft) is formed by two
  intersecting vertical planes making angles of 70º to
  the right and to the left, respectively, to a vertical
  plane passing through the longitudinal axis, as
  viewed when looking aft along the longitudinal axis.

For the anti collision lights:

The aeroplane must have an anticollision
  light system that

(1) Consists of one or more approved
  anti-collision lights located so that their light will
  not impair the crew’s vision or detract from the
  conspicuity of the position lights

(b) Field of coverage. The system must consist
  of enough light to illuminate the vital areas around
  the aeroplane considering the physical configuration
  and flight characteristics of the aeroplane. The field
  of coverage must extend in each direction within at
  least 75º above and 75º below the horizontal plane of
  the aeroplane, except that a solid angle or angles of
  obstructed visibility totalling not more than 0·03
  steradians is allowable within a solid angle equal to
  0·15 steradians centred about the longitudinal axis in
  the rearward direction.

On the aerodynamics effect of vortex generator at the wingtip, I found a paper: Modification of a wing tip vortex by vortex generators. It is an experimental study without a wingtip device though. Citing from the conclusion paragraph:

The results show that substantial redistribution of wake circulation can
  be achieved by the merger of the tip vortex with a co-rotating
  vortex from the VG with the core radius of the merged vortex increased by a factor of up to five relative to the undisturbed
  tip vortex.

The VG was though positioned at a $\eta =\frac{y}{b} = 0.2$ inboard of the wingboard.
Summing up the comments above, and merging this interesting explanation from @Peter Kämpf on winglets, what I think (yes, it is an "opinion", but otherwise I would need to run some 3D CFD and/or a wind-tunnel test) is that:

There would be a vortex generated in certain conditions for sure.
To design this device in order to avoid winglet stall separation would be quite a challenging task. The device its positioned very outboard which would make it be more or less aligned with the winglet. I think that the close span-wise coupling between he device and the winglet would make it hard for the vortex to develop in such a way to re-attach the flow on the winglet.

Vortex generators are located along the wing, not at the wing tip.  They are designed to make the wing stall break over a greater range of angle of attack.  What you show at the wing tip will not have a similar effect as a VG.

Just a comment on this answer. In my opinion is not fully true that VG were never used at the wing tip. See for instance the Gloster Javelin here http://www.aerospaceweb.org/question/aerodynamics/q0009.shtml. These in particular were designed to maintain ailerons effective at high AoA which is another use of VG. The big difference I see with respect to the strobe shield is that this is generally placed at the leading edge whereas those designed to maintain the flow attached at the aileron level are much more downstream.


text:
Am I correctly calculating the density altitude?
I came across the following question while studying for the FAA PPL written test:

Determine the density altitude for these conditions:
Altimeter setting: 30.35
Runway temperature: +25°F
Airport elevation: 3,894 ft. MSL

I performed the following steps:
Temperature (°C) $\color{red}{= 5/9 × 25} = 14 $
Pressure altitude (PA) $= 3,894 + (29.92-30.35) × 930 = 3494 $
Density altitude (DA) $= \text{PA} + 120(14 - 15 + .002×\text{PA}) = 4200 $
Yet that answer is not listed, and according to the DA/PA graph provided, DA is 2000ft.
My final formula seems to be correct according to Wikipedia, plus it also matches my intuition of how PA works: According to the question we have 14°C at 4000ft, as opposed to the expected 7°C at 4000ft; seeing as temperature is higher, DA should be higher than PA.
Where am I going wrong?

Why are you using 930 instead of 1,000 in you calculation of Pressure altitude? https://en.wikipedia.org/wiki/Pressure_altitude. Also, where did the second PA term in the DA formula come from? Not something I use. You might want to check out http://www.flyingmag.com/technique/tip-week/calculating-density-altitude-pencil

@JScarry:  930ft/inHg is the correct factor at about 29.92 — 28.92 inHg; for higher altitudes it increases; thank you for bringing that to my attention, I'll probably use 1000 from now on.  The $.002×\text{PA}$ is to correct for lapse rate.

Check your temperature conversion: 25F is -3.9C...
Should be (25-32)*(5/9).

Yep, I suspected I'd goofed up with something as stupid as a conversion error.  I'd answered several questions about change of density altitude with change of temperature, where you can forget the $-32$, and am embarrassed to say I went and did the same thing here.

I'm studying for the commercial and CFI exams - there are plenty of ways to mess up!  Happy to help.


text:
What help do you have to provide if you are seated in the emergency row and there is a situation?
Follow up from What if there are no volunteers to sit in the emergency exit row? and @mins 's comment:

I'm curious... What prevents someone to say they will assist and then just rush out of the plane when the need to assist is here? Tons of people are ready to promise anything to get what they want, and then just say they didn't understand what they were asked for (or more likely swear that they were not asked anything... show me the evidence...)

..What are actually your responsibilities in case an actual emergency occurs and you have to do the assisting tasks?

You should open the door, if told so by the crew. And I'm pretty sure that this is in your very own interest.

And get out of the way, by evacuating yourself

Some videos showing how to open the overwing exit:
https://www.youtube.com/watch?v=bv9fchRvEig
https://www.youtube.com/watch?v=tg44goIZnqw
https://www.youtube.com/watch?v=2gZ22iQBlmc

Read the evacuation section of this incident report.  Despite being briefed on what to do, passengers seated in an exit row did not open the door and evacuated from another exit on the other side of the aircraft.  Some people are not able to maintain a clear head when under pressure.

When sitting at the exit you need to be the first one out of the plane. Basically when the emergency is declared and the captain says, Evacuate, Evacuate, you go to your door, check the outside for fires. Then you open the door en be the first one out of the door. You then put as much distant between you and the plane as fast as possible. Or you assist on the outside of the plane by making sure people create enough distance between themselves and the plane.
What could with hold them from sitting there is the responsibility to get the door open. Or the fact that they enter a shock and will be unable to move during an emergency.

I would check for any kind of hazards, especially water. I don't want to be the person who sank the plane, without need:   One rear door was opened by a panicking passenger, causing the A320 to fill more quickly with water. https://en.wikipedia.org/wiki/US_Airways_Flight_1549#Evacuation

"What could with hold them from sitting there is [...] Or the fact that they enter a shock and will be unable to move during an emergency." Most people do not know that they will freeze in an emergency, so unless they've gone through something like that before its highly doubtful anybody is going to recognize that trait and not sit in the exit row.

authors: Антон Новичихин